Let a =2 i - j + k and b =λ j +2 k , λ ∈ Z be two vectors. Let c = a × b and d be a vector of magnitude 2 in y z -plane. If | c |= 53 , then the maximum possible value of ( c · d )^ 2 is equal to :

Correct answer: C. 208 a = 2 i - j + k b = λ j + 2 k ; λ ∈ Z c = a × b = (-2 - λ) i - 4 j + 2λ k | c | = 53 5λ^2 + 4λ - 33 = 0 λ = 2.2 or -3 λ = -3 c = i - 4 j - 6 k let d = y j + z k | d | = 2 y^2 + z^2 = 4 ( c · d )^2 = (4y + 6z)^2 ≤ ( 4^2 + 6^2 × y^2 + z^2 )^2 ≤ 208

JEE Main 202622 JanuaryEvening Shift

Exam: JEE Main
Year: 2026
Date: 22 January
Shift: Evening Shift
Subject: Mathematics
Chapter: Vector Algebra

JEE Main 2026 · 22 January · Evening Shift · Mathematics · Vector Algebra

Let a =2 i - j + k and b =λ j +2 k , λ ∈ Z be two vectors. Let c =...

Let

\vec{a}=2 \hat{i}-\hat{j}+\hat{k}

and

\vec{b}=\lambda \hat{j}+2 \hat{k}, \lambda \in \boldsymbol{Z}

be two vectors. Let

\vec{c}=\vec{a} \times \vec{b}

and

\vec{d}

be a vector of magnitude 2 in

y z

-plane. If

|\overrightarrow{\mathrm{c}}|=\sqrt{53}

, then the maximum possible value of

(\overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{d}})^{2}

is equal to :

Options

A26
B52
C208
D104

Correct answer

C. 208

Step-by-step solution

\vec{a} = 2\hat{i} - \hat{j} + \hat{k}

\vec{b} = \lambda\hat{j} + 2\hat{k}

;

\lambda \in Z

\vec{c} = \vec{a} \times \vec{b} = (-2 - \lambda)\hat{i} - 4\hat{j} + 2\lambda\hat{k}

|\vec{c}| = \sqrt{53}

\Rightarrow 5\lambda^2 + 4\lambda - 33 = 0

\lambda = 2.2

-3

\Rightarrow \boxed{\lambda = -3}

\vec{c} = \hat{i} - 4\hat{j} - 6\hat{k}

let

\vec{d} = y\hat{j} + z\hat{k}

|\vec{d}| = 2

\Rightarrow y^2 + z^2 = 4

(\vec{c} \cdot \vec{d})^2 = (4y + 6z)^2 \leq \left(\sqrt{4^2 + 6^2} \times \sqrt{y^2 + z^2}\right)^2 \leq 208

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